Question

What is the quantity of heat required to change 0.02 kg of ice at 0 oC to water at 60 oC ?

[ specific heat capacity of water = 4.2 x 10^3 J kg^-1 o C^-1 ]

[ specific latent heat of fusion of ice = 3.34 x 10^5 J kg^-1 ]

Answer 1

Let's first find the quantity of heat required to convert $0.02\ kg$ of ice at $0^{\circ}C$ to $0.02\ kg$ of water at $0^{\circ}C$:

$$Q_1=mL_f=0.02\ kg\cdot3.34\cdot10^5\ \dfrac{J}{kg}=6680\ J.$$

Then, we can find the quantity of heat required to convert $0.02\ kg$ of water at $0^{\circ}C$ to $0.02\ kg$ of water at $60^{\circ}C$:

Q_2=mc\Delta T=0.02\ kg\cdot4200\ \dfrac{J}{kg\cdot \!^{\circ}C}\cdot60^{\circ}C=5040\ J.

Finally, we can find the total amount of heat required to convert 0.02 kg of ice at $0^{\circ}C$ to water at $60^{\circ}C$:

$$Q=Q_1+Q_2=6680\ J+5040\ J=11720\ J.$$